Number theory problem assignment

Question

$x^2 + x - 3 \equiv 0 \pmod {p^2}$. $p$ is a prime number and satisfies $13^{(p-1)/2} \equiv 1 \pmod p$. I need to find positive value of $x$. Please help me.

Answer 1

If $p=2m+1$ then $x^2+x-3\equiv x^2 -2m-3\equiv (x-m)^2-m^2-3\pmod p$. So your first task is to verify that $m^2+3$ is a square $\pmod p$. With a root $a^2\equiv m^2+3\pmod p$ (so $a^2=m^2+3+kp $ for some $k$), let $x=(m\pm a)+rp$ and compute $$\begin{align} x^2+x-3&=\left((m\pm a)^2+2(m\pm a)rp+r^2p^2\right)+(m\pm a)+rp-3\\ &={(m^2+a^2-3)}\pm (2m+1)a+m+(2m+1\pm a)pr+r^2p^2\\ &=2m^2+kp\pm pa+m+p^2r\pm apr+r^2p^2\\ &=pm+kp\pm pa \pm apr+p^2(r+r^2)\\ &\equiv p(m+k\pm a\pm ar)\end{align}$$ to obtain the condition $r\equiv \mp a^{-1}(m+k)-1$ and hence $$x\equiv m-p\pm (a-a^{-1}(m+k))\pmod{p^2}. $$